Mirrors

Convex Mirror Virtual Image

negative distance, upright scale

A convex mirror gives a negative image distance: the reflected rays spread, while their back-projections meet behind the mirror.

Example

A convex mirror gives a negative image distance: the reflected rays spread, while their back-projections meet behind the mirror. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

A convex mirror has negative focal length

The convex mirror's focal length is -6 metres. With the object 12 metres in front, reflected rays spread instead of meeting in front of the mirror.

f=−6 mu=12 mf = -6\ \text{m}\qquad u = 12\ \text{m}
Convex mirror virtual imageReflected rays spread; back-projections meet.Fmirrorobjectimage

Fixed convex focus keeps images virtual

Hold the convex focal length at -6 metres. Every row keeps a negative image distance, so the image stays behind the mirror while the upright scale changes.

fuvm−6 m6 m−3 m12−6 m12 m−4 m13−6 m18 m−92 m14\begin{array}{c|c|c|c}f&u&v&m\\-6\ \text{m}&6\ \text{m}&-3\ \text{m}&\tfrac{1}{2}\\-6\ \text{m}&12\ \text{m}&-4\ \text{m}&\tfrac{1}{3}\\-6\ \text{m}&18\ \text{m}&\tfrac{-9}{2}\ \text{m}&\tfrac{1}{4}\\\end{array}
Convex mirror virtual imageThe table scans neighboring virtual-image rows.Fmirrorobjectimage

Fixed object distance scans convex focus

Now hold the object distance at 12 metres. Making the focal length more negative moves the virtual image farther behind the mirror.

fuvm−3 m12 m−125 m15−4 m12 m−3 m14−6 m12 m−4 m13\begin{array}{c|c|c|c}f&u&v&m\\-3\ \text{m}&12\ \text{m}&\tfrac{-12}{5}\ \text{m}&\tfrac{1}{5}\\-4\ \text{m}&12\ \text{m}&-3\ \text{m}&\tfrac{1}{4}\\-6\ \text{m}&12\ \text{m}&-4\ \text{m}&\tfrac{1}{3}\\\end{array}

The image reciprocal is negative

Subtracting the object reciprocal from the negative focal reciprocal gives a negative image reciprocal.

1v=1−6 m−112 m=−14 1/m\frac{1}{v} = \frac{1}{-6\ \text{m}} - \frac{1}{12\ \text{m}} = \tfrac{-1}{4}\ 1/\text{m}

Back-projections meet behind the mirror

The checked image distance is -4 metres, so the image is behind the mirror. The positive magnification makes it upright, and the image height is 1 metre.

v=−4 mm=13himage=1 mv = -4\ \text{m}\qquad m = \tfrac{1}{3}\qquad h_{\text{image}} = 1\ \text{m}
Convex mirror virtual imageDashed back-projections meet behind the mirror.Fmirrorobjectimage
optics Negative image distance for a convex mirror marks a virtual image behind the mirror.